Chapter 2
Homework Solutions
Easy
P2.1
P2.5
(a)
v avg 
x 10 m

 5 m/ s
t
2s
(b)
v avg 
5m
 1.2 m/ s
4s
(c)
v avg 
x 2  x 1 5 m  10 m

 2.5 m/ s
t 2  t1
4 s2 s
(d)
v avg 
x 2  x 1 5 m  5 m

 3.3 m/ s
t 2  t1
7 s4s
(e)
v avg 
x 2  x1 0  0

 0 m/ s
t 2  t1 8  0
(a)
At ti = 1.5 s, xi = 8.0 m (Point A)
At tf = 4.0 s, xf = 2.0 m (Point B)
v avg 
x f  xi
t f  ti

(b)

2.0  8.0 m
4.0  1.5 s
6.0 m
 2.4 m/ s
2.5 s
The slope of the tangent line can be found
points C and D. (tC = 1.0 s, xC = 9.5 m) and
3.5 s, xD = 0),
v  3.8 m/ s
(c)
The velocity is zero when x is a minimum.
This is at t  4.0 s .
P2.18
Method One
ANS. FIG. P2.5
from
(tD =
Suppose the unknown acceleration is constant as a car moving at
v i1  35.0 mi h comes to a v f  0 stop in x f 1  x i  40.0 ft . We find its
acceleration from
a
v 2f 1 v i12  2ax f 1  x i :
v 2f 1 v i2
2 x f 1  x i 


0  (35.0 mi h )2 5 280 ft
2  40.0 ft 
mi

2
1h
3 600 s

2
 32.9 ft s 2
v 0
Now consider a car moving at v i2  70.0 mi h and stopping to f
with
a  32.9 ft s 2
. From the same equation its stopping distance is
x f 2  xi 
v 2f 2  v i2
2a

0  70.0 mi/ h 2 5 280 ft

2 32.9 ft s 2 
1 mi

2
1h
3 600 s

2
 160 ft
Method Two
For the process of stopping from the lower speed v i1 we have
v 2f  v i12  2ax f 1  x i  0  v i12  2ax f 1 v i12  2ax f 1
;
;
. For stopping from v i2  2v i1
, similarly
0  v i22  2ax f 2  0 v i22  2ax f 2
;
. Dividing gives
xf2
v i22

2
2
v i1 x f 1 ; x f 2  40 ft  2  160 ft
P2.20
(a)
vi = 100 m/s, a = –5.00 m/s2, vf = vi + at, so 0 = 100 – 5t 
t = 20.0 s
(b)
(c)
No. See reason in (c).
Find the distance (displacement) in which the plane needs to stop:
v 2f  v i2  2a(x f – x i )  v i2  2ax
0  (100)2  2(–5)x  x  1000 m  1.00 km.
The required distance is greater than 0.800 km. The plane would
overshoot the runway.
P2.25
(a)
The time it takes the truck to reach 20.0 m/s is found from v f  v i  at.
Solving for t yields
t
v f  vi
a

20.0 m s  0 m s
 10.0 s
2.00 m s 2
The total time is thus 10.0 s  20.0 s  5.00 s  35.0 s
(b)
The average velocity is the total distance traveled divided by the total
time taken. The distance traveled during the first 10.0 s is
x 1  vt 


0  20.0
10.0  100 m
2
With a being 0 for this interval, the distance traveled during the next 20.0
s is
x 2  v it 
1 2
at  20.020.0  0  400 m
2
The distance traveled in the last 5.00 s is
x 3  vt 


20.0  0
5.00  50.0 m
2
The total distance x  x 1  x 2  x 3  100  400  50  550 m, and the
x 550
average velocity is given by v  
 15.7 m s .
t 35.0
P2.30
(a)
We know vf = 0 when t = 3.00 s, and a = −g = −9.80 m/s2.
v f  v i  at
0  v i  gt  v i  gt
v i  9.80 m/ s 2 3.00 s   29.4 m/ s
(b)
Find the vertical displacement ∆y:
y  y f  y i 


1
vi  v f t
2
1
29.4 m/ s  03.00 s 
2
y  44.1 m
y 
P2.32
(a)
Consider the upward flight of the arrow.
v yf2  v yi2  2ay y f  y i 
0  100 m s   2 9.8 m s 2 y
2
y 
(b)
10 000 m 2 s 2
 510 m
19.6 m s 2
Consider the whole flight of the arrow.
y f  y i  v yit 
1 2
at
2 y
0  0  100 m s t 
1
9.8 m s 2 t 2

2
The root t = 0 refers to the starting point. The time of flight is given by
t
P2.41
100 m s
 20.4 s
4.9 m s 2
v f  vi
(a)
a
(b)
x f  v it 
t

632 5 280 / 3 600
 662 ft/ s 2  202 m/ s 2
1.40
 5 280 ft/ mi 
1 2
at  632 mi/ h 
1.40 s 
2
 3 600 s/ h 
1
2
662 ft/ s 2 1.40 s 

2
 649 ft  198 m

Medium
P2.4
(a)
Let d represent the distance between A and B. Let t1 be the time for
d
which the walker has the higher speed in 5.00 m/ s  . Let t2 represent
t1
d
the longer time for the return trip in 3.00 m/ s   . Then the times are
t2
d
d
t1 
and t 2 
. The average speed is:
5.00 m/ s 
3.00 m/ s 
Total d istance
Total time
dd
2d


d / 5.00 m/ s   d / 3.00 m/ s  8.00 m/ s d / 15.0 m 2 / s 2 
v avg 
v avg 
2 15.0 m 2 / s 2 
8.00 m/ s
 3.75 m/ s
P2.7
P2.9
(b)
She starts and finishes at the same point A. With total displacement = 0,
average velocity = 0 .
(a)
v
5  0 m
1  0 s
(b)
v
5  10 m
4  2  s
(c)
v
5  5 m 
5 s  4 s 
(d)
v
0  5 m 
 5 m/ s
8 s  7 s 
(a)
The tortoise crawls through a distance D before the rabbit resumes the
race. When the rabbit resumes the race, the rabbit must run through 200
m at 8.00 m/s while the tortoise crawls through the distance (1 000 m –
D) at 0.200 m/s. Each takes the same time interval to finish the race:
 5 m/ s
 2.5 m/ s
0
ANS. FIG. P2.7
 200 m   1 000 m  D 
t  

 8.00 m/ s   0.200 m/ s 
 0.200 m/ s 200 m   8.00 m/ s 1 000 m  D 
1 000 m  D 
0.200 m/ s 200 m 
8.00 m/ s
 D  995 m
So, the tortoise is 1 000 m – D = 5.00 m from the finish line when the
rabbit resumes running.
(b)
Both begin the race at the same time: t = 0. The rabbit reaches the 800-m
position at time t = 800 m/(8.00 m/s) = 100 s. The tortoise has crawled
through 995 m when t = 995 m/(0.200 m/s) = 4 975 s. The rabbit has
waited for the time interval ∆t = 4 975 s – 100 s = 4 875 s .
P2.10
(a)
At t = 2.00 s, x = [3.00(2.00)2 – 2.00(2.00) + 3.00] m = 11.0 m.
At t = 3.00 s, x = [3.00(3.00)2 – 2.00(3.00) + 3.00] m = 24.0 m
so
v avg 
(b)
x 24.0 m  11.0 m

 13.0 m/ s
t
3.00 s  2.00 s
At all times the instantaneous velocity is
v
d
3.00t 2  2.00t  3.00 6.00t  2.00 m/ s

dt
At t = 2.00 s, v = [6.00(2.00) – 2.00] m/s = 10.0 m/ s
At t = 3.00 s, v = [6.00(3.00) – 2.00] m/s = 16.0 m/ s
(c)
aavg 
v 16.0 m/ s  10.0 m/ s

 6.00 m/ s 2
t
3.00 s  2.00 s
(d) At all times a 
d
6.00t  2.00  6.00 m/ s 2 . This includes both t = 2.00 s
dt
and t = 3.00 s.
(e)
P2.11
From (b), v = (6.00t – 2.00) = 0  t = (2.00)/(6.00) = 0.333 s.
x = 2.00 + 3.00t – t2, so v 
dx
dv
 3.00  2.00t, and a 
 2.00
dt
dt
At t = 3.00 s:
P2.26
(a)
x = (2.00 + 9.00 – 9.00) m = 2.00 m
(b)
v = (3.00 – 6.00) m/s = –3.00 m/ s
(c)
2
a = –2.00 m/ s
(a)
Compare the position equation x = 2.00+ 3.00t – 4.00t2 to the general form
x f  x i  v it 
1 2
at
2
to recognize that xi = 2.00 m, vi = 3.00 m/s, and a = –8.00 m/s2. The velocity
equation, vf = vi + at, is then
vf = 3.00 m/s – (8.00 m/s2)t
The particle changes direction when vf = 0, which occurs at t 
position at this time is
2
3 
3 
x  2.00 m  3.00 m/ s  s   4.00 m/ s 2  s   2.56 m
8 
8 
P2.33
(a)
We know when t = 1.50 s, ∆y = (yf – yi) = 4.00 m; also, a = –g.
1 2
gt
2
2
4.00 m  v i 1.50 s   4.90 m/ s 2 1.50 s 
y f  y i  v it 
v i  10.0 m/ s up
(b)
Solve for final velocity, vf :
v f  v i  at
v f  10.0 m/ s  9.80 m/ s 2 1.50 s 
v f  4.7 m/ s
The final velocity is 4.7 m/ s, down.
P2.35
We have
1 2
gt
2
0  30 m  8.00 m/ s t  4.90 m/ s 2 t 2
y f  y i  v it 
Solving for t,
t
8.00 
t  1.79 s
8.002  4 4.9030 8.00  64  588

2 4.90
9.80
3
s. The
8
Hard
P2.51
Consider the runners in general. Each completes the race in a total time
interval T. Each runs at constant acceleration a for a time interval ∆t, so each
covers a distance (displacement) ∆xa = ½a∆t2 where they eventually reach a
final speed (velocity) v = a∆t, after which they run at this constant speed for
the remaining time (T – ∆t) until the end of the race, covering distance ∆xv =
v(T – ∆t) = a∆t (T – ∆t). The total distance (displacement) each covers is the
same:
∆x = ∆xa + ∆xv
∆x = ½a∆t2 + a∆t (T – ∆t)
∆x = a[½∆t2 + ∆t (T – ∆t)]
so
a
x
1 2
t  t T  t 
2
where ∆x = 100 m, and T = 10.4 s.
(a)
For Laura (runner 1), ∆t1 = 2.00 s:
2
a1 = (100 m)/(18.8 s2) = 5.32 m/ s
For Healan (runner 2), ∆t2 = 3.00 s:
2
a2 = (100 m)/(26.7 s2) = 3.75 m/ s
(b)
Laura (runner 1): v1 = a1∆t1 = 10.6 m/ s
Healan (runner 2): v2 = a2∆t2 = 11.2 m/ s
(c)
The 6.00-s mark occurs after either time interval ∆t. From the reasoning
above, each has covered the distance
∆x = a[½∆t2 + ∆t(t – ∆t)]
where t = 6.00 s.
Laura (runner 1): ∆x1 = 53.19 m
Healan (runner 2): ∆x2 = 50.56 m
So, Laura is ahead by (53.19 m  50.56 m)  2.63 m.
(d) Laura accelerates at the greater rate, so she will be ahead of Healen at,
and immediately after, the 2.00-s mark. After the 3.00-s mark, Healan is
travelling faster than Laura, so the distance between them will shrink. In
the time interval
from the 2.00-s mark to the 3.00-s mark, the distance between
them will be the greatest.
During that time interval, the distance between them (the position of
Laura relative to Healan) is
D = ∆x1 – ∆x2 = a1[½∆t12 + ∆t1(t – ∆t1)] – ½a2t2
because Laura has ceased to accelerate but Healan is still accelerating.
Differentiating with respect to time, (and doing some simplification), we
can solve for the time t when D is an maximum:
dD/dt = a1∆t1
t – a2t = 0 
t = ∆t1(a1/a2) = 2.84 s.
Substituting this time back into the expression for D, we find that D =
4.47 m, that is, Laura ahead of Healan by 4.47 m.
P2.55
The rock falls a distance d for a time interval ∆t1 and the sound of the splash
travels upward through the same distance d for a time interval ∆t2 before the
man hears it. The total time interval ∆t = ∆t1 + ∆t2 = 2.40 s.
(a)
Relationship between distance the rock falls and time interval ∆t1: d = ½
g ∆t12
Relationship between distance the sound travels and time interval ∆t2: d
= vs∆t2, where vs = 336 m/s.
d  v s t 2 
1
gt 12
2
Substituting t1  t  t 2 gives
2
v s t 2
2
 t  t 2 
g

t 2 2  2  t 

vs 
t 2  t 2  0
g 
t 2 2  2  2.40 s 
336 m/ s 
2
t 2  2.40 s   0
2
9.80 m/ s 
t 2 2  73.37 t 2  5.76  0
Solving the quadratic equation gives
∆t2 = 0.078 6 s  d = vs∆t2 = 26.4 m
(b)
Ignoring the sound travel time, d 
error of 6.82% .
1
2
9.80 m/ s 2 2.40 s   28.2 m, an

2